Question

I really need help with Algebra 2 Quadratic Relations and Conic Functions!

Identify the conic section, any lines of symmetry, and the domain and range.

x=2y^2+5

I need help understand how to do this!

I will fan anyone who can help me! 😰

Answer 1

umm... I'm not sure. I'll fan you to make up for my failure to help you. :P

Answer 2

@tom982

Answer 3

Firstly, do you know how to identify a conic section?

Answer 4

A circle is when x^2 and y^2 have the same coefficients, an ellipse is when it's the same sign but different coefficients, and a hyperbola is when it's different signs.

Answer 5

@tom982

Answer 6

@tom982 please help me :(

Answer 7

@CGGURUMANJUNATH

Answer 8

Is that meant to be x^2 in the original equation then? You're talking about x^2 when it's not relevant here.

Answer 9

It says the equation is "x=2y^2+5" and I need to find the conic section, lines of symmetry, and the domain and range @tom982

Answer 10

Okay. Then you're barking up the wrong tree with x^2. Read this:

http://www.ck12.org/book/CK-12-Algebra-II-with-Trigonometry-Concepts/section/10.11/

Can you classify it now?

Answer 11

Hyperbola?

Answer 12

A conic is of the form \(Ax^2+Bxy+Cy^2+Dx+Ey+F=0\). Rearranging yours into this form we get: \(2y^2-x+5=0\). This gives us our constants \(A=0\), \(B=0\), \(C=2\)\, \(D=-1\), \(E=0\) and \(F=5\). The discriminant is \(B^2-4AC\) and in this case it is \(0^2-4\times0\times2=0\). As the discriminant is zero, what does this say about our conic? Refer to the page I linked you to.

Answer 13

Ah so a parabola

Answer 14

Exactly, well done. Parabolas are symmetric about the x-axis whenever (x,y)=(x,-y) which we can easily check: \(x=2(-y)^2+5=2y^2+5\) as required. Hence the x-axis (the line y=0) is the only line of symmetry for this parabola.

Can you find the domain and range?

Answer 15

Domain: -5<=x<=5
Range: -5<=y<=5?

Answer 16

Why have you restricted the domain? What's wrong with using -6?
\(x=2(-6)^2+5=77\)

Answer 17

Wait how did you get that???

Answer 18

Ah sorry I misread what you said, but you're still wrong. Your function is written abnormally: \(f(y)=x\) so our domain is for y and the range is for x. There are no numbers we can't put into this equation so the domain is \(y \in \mathbb{R} \). What numbers can we get out of it?